Page 144 - Calculus Demystified

CHAPTER 5
SOLUTION Indeterminate Forms 131
The expression is indeterminate of the form ∞−∞. We put the two fractions
over a common denominator to obtain
e 4x − 1 − 4x
lim .
x→0 4x(e 4x − 1)
Notice that the numerator and denominator both tend to zero as x → 0, so this
is indeterminate of the form 0/0. Therefore l’Hôpital’s Rule applies and our
limit equals
4e 4x − 4
lim .
4x
x→0 4e (1 + 4x) − 4
Again the numerator and denominator tend to zero and we apply l’Hôpital’s
Rule; the limit equals
16e 4x 1
lim = .
4x
x→0 16e (2 + 4x) 2

1 2
You Try It: Evaluate the limit lim x→0 + 2 .
cos x − 1 x
5.2.5 OTHER ALGEBRAIC MANIPULATIONS

Sometimes a factorization helps to clarify a subtle limit:

EXAMPLE 5.12
Evaluate the limit

2 4 2 1/2
lim x − (x + 4x + 5) .
x→+∞
SOLUTION
The limit as written is of the form ∞−∞. We rewrite it as

1 − (1 + 4x −2 + 5x −4 1/2
)

2
lim x 1 − (1 + 4x −2 + 5x −4 1/2 = lim .
)
x→+∞ x→+∞ x −2
Notice that both the numerator and denominator tend to zero, so it is now
indeterminate of the form 0/0. We may thus apply l’Hôpital’s Rule. The result
is that the limit equals
)
(−1/2)(1 + 4x −2 + 5x −4 −1/2 · (−8x −3 − 20x −5 )
lim
x→+∞ −2x −3
−2 −4 −1/2 −2
= lim −(1 + 4x + 5x ) · (2 + 5x ).
x→+∞